Properties of addition on rational numbers are similar to properties of addition on integers.
Let \(\frac{a}{b}\) and \(\frac{c}{d}\), be any two unique rational numbers.
Property 1:
Closure Property: The addition of any two rational numbers is a rational number.
Therefore, \(\frac{a}{b}\) + \(\frac{c}{d}\) is a rational number.
Verification:
We have,
1) \(\frac{3}{4}\) + \(\frac{5}{7}\) = \(\frac{(3 X 7) + (5 X 4)}{28}\)
= \(\frac{(21 + 20)}{28}\) = \(\frac{41}{28}\)
which is a rational number.
2) \(\frac{-7}{12}\) + \(\frac{5}{6}\) = \(\frac{(-7) + (5 X 2)}{12}\)
= \(\frac{(-7 + 10)}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
which is a rational number.
Property 2:
Commutativity: Two rational numbers can be added in any order.
Therefore, \(\frac{a}{b}\) + \(\frac{c}{d}\) = \(\frac{c}{d}\) + \(\frac{a}{b}\).
Verification:
We have,
1) \(\frac{5}{2}\) + \(\frac{4}{7}\) = \(\frac{(5 X 7) + (4 X 2)}{14}\)
= \(\frac{(35 + 8)}{14}\) = \(\frac{43}{14}\),
\(\frac{4}{7}\) + \(\frac{5}{2}\) = \(\frac{(4 X 2) + (5 X 7)}{14}\)
= \(\frac{(8 + 35)}{14}\) = \(\frac{43}{14}\)
Therefore, \(\frac{5}{2}\) + \(\frac{4}{7}\) = \(\frac{4}{7}\) + \(\frac{5}{2}\)
2) \(\frac{13}{5}\) + \(\frac{-8}{11}\) = \(\frac{(13 X 11) + (-8 X 5)}{55}\)
= \(\frac{(143 + (-40))}{55}\) = \(\frac{103}{55}\),
\(\frac{-8}{11}\) + \(\frac{13}{5}\) = \(\frac{(-8 X 5) + (13 X 11)}{55}\)
= \(\frac{((-40) + 143)}{55}\) = \(\frac{103}{14}\)
Therefore, \(\frac{13}{5}\) + \(\frac{-8}{11}\) = \(\frac{-8}{11}\) + \(\frac{13}{5}\)
Property 3:
Associativity: For any three raiona1 numbers \(\frac{a}{b}\), \(\frac{c}{d}\), \(\frac{e}{f}\), we have ( \(\frac{a}{b}\) + \(\frac{c}{d}\) ) + \(\frac{e}{f}\) = \(\frac{a}{b}\) + ( \(\frac{c}{d}\) + \(\frac{e}{f}\) ) i.e., while adding three or more rational numbers, they can be grouped in any order.
Verification:
Let \(\frac{4}{7}\), \(\frac{-5}{12}\), \(\frac{9}{-11}\) be three rational numbers. Then,
( \(\frac{4}{7}\) + \(\frac{-5}{12}\) ) + \(\frac{9}{-11}\) = \(\frac{48 – 35}{84}\) + \(\frac{9}{-11}\)
= \(\frac{13}{84}\) + \(\frac{-9}{11}\) = \(\frac{143 – 756}{924}\)
= \(\frac{-613}{924}\)
\(\frac{4}{7}\) + ( \(\frac{-5}{12}\) + \(\frac{9}{-11}\) ) = \(\frac{4}{7}\) + \(\frac{-55 – 108}{132}\)
= \(\frac{4}{7}\) + \(\frac{-163}{132}\) = \(\frac{528 – 1141}{924}\)
= \(\frac{-613}{924}\)
Therefore, ( \(\frac{4}{7}\) + \(\frac{-5}{12}\) ) + \(\frac{9}{-11}\) = \(\frac{4}{7}\) + ( \(\frac{-5}{12}\) + \(\frac{9}{-11}\) ).
This property enables us to add three or more rational numbers by re-arranging and grouping them in a convenient form so that the sum may be found more easily.
Example 1:
Simplify \(\frac{2}{5}\) + \(\frac{8}{3}\) + \(\frac{-11}{15}\) + \(\frac{4}{5}\) + \(\frac{-2}{3}\).
Solution:
By grouping pairs of numbers with common denominator.
\(\frac{2}{5}\) + \(\frac{8}{3}\) + \(\frac{-11}{15}\) + \(\frac{4}{5}\) + \(\frac{-2}{3}\) = ( \(\frac{2}{5}\) + \(\frac{4}{5}\) ) + ( \(\frac{8}{3}\) + \(\frac{-2}{3}\) ) + \(\frac{-11}{15}\)
= \(\frac{2 + 4}{5}\) + \(\frac{8 – 2}{3}\) + \(\frac{-11}{15}\) = \(\frac{6}{5}\) + \(\frac{6}{3}\) + \(\frac{-11}{15}\)
= \(\frac{( 6 x 3 + 6 x 5 + 1 x (-11))}{15}\) = \(\frac{( 18 + 30 – 11)}{15}\)
= \(\frac{37}{15}\)
Example 2:
Simplify \(\frac{4}{3}\) + \(\frac{3}{5}\) + \(\frac{-2}{3}\) + \(\frac{-11}{5}\).
Solution:
By grouping pairs of numbers with common denominator.
\(\frac{4}{3}\) + \(\frac{3}{5}\) + \(\frac{-2}{3}\) + \(\frac{-11}{5}\) = ( \(\frac{4}{3}\) + \(\frac{3}{5}\) ) + ( \(\frac{-2}{3}\) + \(\frac{-11}{5}\) )
= \(\frac{4 + (-2)}{3}\) + \(\frac{3 + (-11)}{5}\) = \(\frac{2}{3}\) + \(\frac{-8}{5}\)
= \(\frac{( 10 + (-24))}{15}\) = \(\frac{-14}{15}\)
Property 4:
Property of Zero: The sum of any rational number and zero is the rational number itself, i.e., if \(\frac{a}{b}\) is any rational number, then \(\frac{a}{b}\) + 0 = 0 + \(\frac{a}{b}\) = \(\frac{a}{b}\).
Verification:
We have,
1) \(\frac{7}{8}\) + 0 = \(\frac{7}{8}\) + \(\frac{0}{8}\)
= \(\frac{7 + 0}{8}\) = \(\frac{7}{8}\)
which is the same rational number.
2) 0 + \(\frac{17}{-11}\) = \(\frac{0}{11}\) + \(\frac{-17}{11}\)
= \(\frac{0 – 17}{11}\) = \(\frac{-17}{11}\)
which is the same rational number.
Example:
Simplify \(\frac{4}{7}\) + \(\frac{-8}{9}\) + \(\frac{-13}{7}\) + 0 + \(\frac{17}{21}\).
Solution:
\(\frac{4}{7}\) + \(\frac{-8}{9}\) + \(\frac{-13}{7}\) + 0 + \(\frac{17}{21}\) = \(\frac{4}{7}\) + \(\frac{-8}{9}\) + ( \(\frac{-13}{7}\) + 0 ) + \(\frac{17}{21}\) = \(\frac{4}{7}\) + \(\frac{-8}{9}\) + \(\frac{-13}{7}\) + \(\frac{17}{21}\)
By grouping pairs of numbers with common denominator.
( \(\frac{4}{7}\) + \(\frac{-13}{7}\) ) + \(\frac{-8}{9}\) + \(\frac{17}{21}\)
= \(\frac{-9}{7}\) + \(\frac{-8}{9}\) + \(\frac{17}{21}\) = \(\frac{-9 x 9 + (-8) x 7 + 17 x 3}{63}\)
= \(\frac{-81 – 56 + 51}{63}\) = \(\frac{-137 + 51}{63}\) = \(\frac{-86}{63}\)
0 is called the additive identity or the identity element for the addition of rational numbers.
Property 5:
Existence of Negative ( Additive inverse ) of a Rational Number: If \(\frac{a}{b}\) is a rational number, then if \(\frac{-a}{b}\) is a rational number such that \(\frac{a}{b}\) + ( \(\frac{-a}{b}\) ) = ( \(\frac{-a}{b}\) ) + \(\frac{a}{b}\) = 0.
\(\frac{-a}{b}\) is called negative of \(\frac{a}{b}\). It is also called the additive inverse of \(\frac{a}{b}\).
If \(\frac{-a}{b}\) is the negative of \(\frac{a}{b}\) then, \(\frac{a}{b}\) is the negative of \(\frac{-a}{b}\), i.e., \(\frac{a}{b}\) = -( \(\frac{-a}{b}\) )
We have, 0 + 0 = 0 = 0 + 0, so 0 is the additive inverse of itself.
0 is the only rational number which is its own additive inverse.
Verification:
1) \(\frac{7}{8}\)
Additive inverse of \(\frac{7}{8}\) is -( \(\frac{7}{8}\) )
= \(\frac{-7}{8}\)
We have,
\(\frac{7}{8}\) + \(\frac{-7}{8}\) = \(\frac{7 – 7}{8}\)
= \(\frac{0}{8}\) = 0
\(\frac{-7}{8}\) + \(\frac{7}{8}\) = \(\frac{-7 + 7}{8}\)
= \(\frac{0}{8}\) = 0
Therefore, \(\frac{7}{8}\) + \(\frac{-7}{8}\) = \(\frac{-7}{8}\) + \(\frac{7}{8}\) = 0.
2) \(\frac{-14}{19}\)
Additive inverse of \(\frac{-14}{19}\) is -( \(\frac{-14}{19}\) )
= \(\frac{14}{19}\)
We have,
\(\frac{-14}{19}\) + \(\frac{14}{19}\) = \(\frac{-14 + 14}{19}\)
= \(\frac{0}{19}\) = 0
\(\frac{14}{19}\) + \(\frac{-14}{19}\) = \(\frac{14 – 14}{19}\)
= \(\frac{0}{19}\) = 0
Therefore, \(\frac{-14}{19}\) + \(\frac{14}{19}\) = \(\frac{14}{19}\) + \(\frac{-14}{19}\) = 0.
2) \(\frac{9}{-16}\)
Additive inverse of \(\frac{9}{-16}\) is -( \(\frac{9}{-16}\) )
= -( \(\frac{-9}{16}\) ) = \(\frac{9}{16}\)
We have,
\(\frac{9}{-16}\) + \(\frac{9}{16}\) = \(\frac{-9}{16}\) + \(\frac{9}{16}\)
= \(\frac{-9 + 9}{16}\) = \(\frac{0}{16}\) = 0
\(\frac{9}{16}\) + \(\frac{9}{-16}\) = \(\frac{9}{16}\) + \(\frac{-9}{16}\)
= \(\frac{9 – 9}{16}\) = \(\frac{0}{16}\) = 0
Therefore, \(\frac{9}{-16}\) + \(\frac{9}{16}\) = \(\frac{9}{16}\) + \(\frac{9}{-16}\) = 0.